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  4. The modulus of the complex number z such that |z+3-i|=1 and arg (z)=π is equal to

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Q. The modulus of the complex number $z$ such that $|z+3-i|=1$ and $\arg (z)=\pi$ is equal to

Complex Numbers and Quadratic Equations

Solution:

Let $z=x+i y$
$ \therefore|z+3-i|=|(x+3)+i(y-1)|=1 $
$ \Rightarrow \sqrt{(x+3)^2+(y-1)^2}=1 $...(i)
$ \because \arg z=\pi$
$ \Rightarrow \tan ^{-1} \frac{y}{x}=\pi $
$ \Rightarrow \frac{y}{x}=\tan \pi=0$
$ \Rightarrow y=0 $...(ii)
From Eqs. (i) and (ii), we get
$ x=-3, y =0$
$\therefore z =-3$
$\Rightarrow |z| =|-3|=3$